Mathematics 376 – Mathematical Statistics Solutions for Final Examination May 16, 2012

I. Let X_{1}, X_{2}, X_{3}, X_{4}, X_{5} be a random sample from a normal distribution with mean μ = 6 and variance σ^{2} = 81.

A) What is the distribution (give the type and all relevant parameter values) of U = \frac{(X_{1}−6)^{2}+(X_{2}−6)^{2}+(X_{3}−6)^{2}+(X_{4}−6)^{2}}{81} ? Why?

Solution: For each i, \frac{X_{i}−6}{9} has a standard normal distribution. Therefore

U = (\frac{X_{1} − 6}{9})^{2} + (\frac{X_{2} − 6}{9})^{2} + (\frac{X_{3} − 6}{9})^{2} + (\frac{X_{4} − 6}{9})^{2}

is the sum of the squares of four independent standard normals. So U has a χ^{2} distribution with 4 degrees of freedom.

B) What is the distribution (give the type and all relevant parameter values) of V = \frac{2(X_{5}−6)}{9\sqrt{U}}? Why?

Solution: We have

V = \frac{\frac{X_{5}−6}{9}}{\sqrt{U/4}}

This has the form of a standard normal, divided by the square root of a χ^{2}, divided by its number of degrees of freedom. So, by definition, V has a t-distribution with 4 degrees of freedom.

C) (10) What is the distribution (give the type and all relevant parameter values) of W = (X_{1} − 6)^{2}/(X_{2} − 6)^{2}? Why?

Solution: We have

W = \begin{matrix}(\frac{X_{1}−6}{9}) & 2 \\ (\frac{X_{2}−6}{9}) & 2 \end{matrix}

This is a ratio of χ^{2}(1) random variables. Hence W has an F -distribution with 1 degree of freedom in the numerator and 1 degree of freedom in the denominator.

II. A random variable Y has pdf of the form

f(y) = \{ \begin{matrix}\frac{2y}{α} e^{−y^{2}/α} & if y ≥ 0 \\ 0 & otherwise \end{matrix}

(This is a particular case of a Weibull distribution; these are often used to model length of life of mechanical components and living organisms.) Let Y_{1}, …, Y_{n} be a random sample from this distribution with unknown α.

A) Find the maximum-likelihood estimator for α.

Solution: The likelihood function is

L(y_{1}, …, y_{n} | α) = \frac{2^{n} ∏_{i} y_{i}}{α^{n}} e^{−(∑_{i} y_{i}^{2})/α}

so

ln(L) = ln(2^{n} ∏_{i} y_{i}) − n ln(α) − \frac{∑_{i} y_{i}^{2}}{α} .

First, let us see if ln(L) has any critical points as a function of α:

0 = \frac{d}{dα} ln(L) = \frac{−n}{α} + \frac{∑_{i} y_{i}^{2}}{α^{2}}

when

α = \frac{1}{n} ∑_{i=1}^{n} y_{i}^{2}

It is easy to check that the second derivative is negative here, so the maximumlikelihood estimator for α is

α̂ = \frac{1}{n} ∑_{i=1}^{n} y_{i}^{2}.

(Note that this is the second sample moment. In fact the variance of Y is a constant multiple of α.)

B) What is the pdf for the sample maximum, Y_{(n)}?

By integration (substitution with u = y^{2}), we see the cdf for this distribution is

F (y) = −e^{−y^{2}/α} + c

for some constant of integration c. At y = 0 we should get F (0) = 0, so c = 1. Hence by the usual formula, the pdf for the sample maximum is

f_{n}(y) = n (1 − e^{−y^{2}/α})^{n−1} \frac{2y}{α} e^{−y^{2}/α}.

III. Let p be the proportion of letters mailed in Belgium that are delivered the next day.

A) A random sample of n = 400 letters is selected for tracking and 373 of those are delivered the next day. Find an approximate 98% confidence interval for p based on this sample.

Solution: We have p̂ = 373/400 = .9325. Since 400 > 30, to get a 98% confidence interval we can use the large sample two-sided formulas with α = .01. The confidence interval is

\begin{aligned}p̂± z_{.01}\sqrt{\frac{p̂(1 − p̂)}{400}} & ≐ .9327 ± 2.33 · .0125 \\ & ≐ (.9033, .9617) \end{aligned}

All values in this interval are thought of as “reasonable estimates” for the proportion p. Since p = .85 is not contained, the answer to the last part is no.

B) (15) (“Thought question”) Note that part A says “approximate.” What is the actual distribution of Y = the number of letters delivered the next day (out of a random sample of size n = 400)? Why does the method you used in part A give a reasonable interval estimate for p? Explain.

Solution: Y has a binomial distribution with n = 400, and p = the actual proportion of letters delivered the next day. An “exact” confidence interval would use information about that binomial distribution to develop the locations of the endpoints. The method we used in part A depends on the properties of the standard normal distribution. That is reasonable here since the Central Limit Theorem implies that \frac{Y}{400} has an approximately normal distribution. (Recall that the estimator \frac{Y}{400} for p can be thought of as a sample mean for 400 independent Bernoulli trials, all with success probability p, so the CLT applies in the form we discussed last semester.)

IV. The fill weights of a random sample of n_{1} = 13 boxes of Twigs and Nuts Extra Crunchy Granola produced at Plant 1 had a mean of 20.5 ounces and standard deviation s = .1 ounce. A similar sample of size n_{2} = 11 produced at Plant 2 had mean fill weight 20.3 ounces and standard deviation s = .083 ounce.

A) (20) Is there sufficient evidence at the α = .01 level to conclude that the variances of the fill weights at the two plants are different?

Solution: To test for equality of variances, we want to use an F -test. The null hypothesis is H_{0} : σ_{1}^{2} = σ_{2}^{2} = σ^{2} and the alternative is H_{a} : σ_{1}^{2} ≠ σ_{2}^{2}. Under H_{0}, the test statistic

F = S_{1}^{2}/S_{2}^{2} (which equals \frac{(12S_{1}^{2}/σ^{2})/12}{(10S_{2}^{2}/σ^{2})/10} )

has an F -distribution with 12 degrees of freedom in the numerator and 10 degrees of freedom in the denominator. For a test with α = .01, we want to set the rejection region to be

\{F | F > f_{.005}(12, 10)\} ∪ \{F | 0 < F < f_{.995}(12, 10) = \frac{1}{f_{.005}(10, 12)}\},

which works out to the union

(5.66, +∞) ∪ (0, .1964)

Our test statistic value is

F = (.1)^{2}/(.083)^{2} ≐ 1.45

This is not in the rejection region, so we do not reject H_{0}.

B) (20) Is there sufficient evidence to conclude that the mean fill weights are different? Report the results by giving an estimate of the p-value of your test.

Solution: Because we did not find evidence to conclude the variances were different, we will use the basic small-sample t-test for equality of means (two-tailed version). The pooled estimator for the variance is

S^{2} ≐ \frac{(12)(.1)^{2} + (10)(.083)^{2}}{22} = .008586

This gives S ≐ .0927. Our test statistic is

t = \frac{20.5 − 20.3}{.0927\sqrt{1/13 + 1/11}} ≐ 5.27

From the t-table, for 22 degrees of freedom, we have t_{.005} = 2.819. We can say p < 2 · (.005) = .01. So there is relatively strong evidence to conclude that the mean fill weights are different. (Using R, in fact the p value is quite a bit smaller even than that: p ≐ .000028.)

V. The following table gives measurements of water depth remaining in an evaporating reservoir as a function of time:

x time (weeks) y depth (meters)
1  19.8
4  16.5
14  12.8
32  8.1
52  7.5

A) (20) Find the least-squares estimators for the coefficients β_{0}, β_{1} in a model Y = β_{0} + β_{1}x + ε for this data set.

Solution: Organizing the calculation the way we discussed in class,

\overline{x} = 20.6
S_{xx} = 1819.2
\overline{y} = 12.94
S_{xy} = −418.62
β̂_{1} = \frac{S_{xy}}{S_{xx}} ≐ −0.23
β̂_{0} = \overline{y} − β̂_{1}\overline{x}
≐ 17.68

So the estimated model would be

Y = 17.68 − .23x + ε

B) (20) Is there sufficient evidence to say that β_{1} < −0.1? Test with α = .05.

Solution: This is a lower-tail t-test with a null hypothesis saying β_{1} = −0.1 (or something larger). The test statistic is

t = \frac{β̂_{1} − (−.1)}{S\sqrt{c_{11}}}

So we need to compute additionally

\begin{aligned}S_{yy} & = 112.772 \\ S^{2} & = \frac{1}{5 − 2} (S_{yy} − β̂_{1}S_{xy}) ≐ 5.48 \\ S & = \sqrt{5.48} ≐ 2.34118 \\ c_{11} & = \frac{1}{S_{xx}} ≐ .0005497 \end{aligned}

So then

t = \frac{(−.23) + .1}{(2.34118)\sqrt{.0005497}} ≐ −2.368

For n−2 = 3 degrees of freedom, −t_{.05} = −2.353. We are (just barely!) in the rejection region, so we would say that there is enough evidence to conclude β_{1} < −0.1.

VI. (20) Suppose we have measurements taken from a normal population with unknown μ and known σ^{2}. We test H_{0} : μ = μ_{0} versus H_{a} : μ > μ_{0} using a Z-test and a rejection region designed to produce a Type I error probability α. Show that in order to make the Type II error probability equal a given β when μ is equal to some μ_{a} > μ_{0}, the sample size should be selected as follows:

n ≥ \frac{(z_{α} + z_{β})^{2}σ^{2}}{(μ_{a} − μ_{0})^{2}} .

Solution: To get a given Type I error probability, α, we make the rejection region for the upper tail test

RR = \{Z | Z > z_{α}\} ↔ \{\overline{Y} > μ_{0} + z_{α} · \frac{σ}{\sqrt{n}}\}

Now, assuming the actual population mean is μ_{a}, to compute the Type II error probability, we consider the case where we would fail to reject H_{0}, but where H_{0} is actually false. That is we want the following:

\begin{aligned}β & = P (\overline{Y} ≤ μ_{0} + z_{α} · \frac{σ}{\sqrt{n}}) \\ & = P (\frac{\overline{Y} − μ_{a}}{σ/\sqrt{n}} ≤ \frac{(μ_{0} − μ_{a})}{σ/\sqrt{n}} + z_{α}) \end{aligned}

Since we have “restandardized” in the last equation, to get the probability to equal β, the right side of the inequality should be

\frac{(μ_{0} − μ_{a})}{σ/\sqrt{n}} + z_{α} = −z_{β}

This rearranges to

\sqrt{n} = \frac{(z_{α} + z_{β})σ}{(μ_{a} − μ_{0})}

and we get the desired equality after squaring.

Extra Credit

A New York Times/CBS poll conducted between April 5 and April 12, 2010 included the question “Do you approve or disapprove of the way your Representative in Congress is handling his or her job?” The poll was carried out in two stages. First a large national simple random sample was asked the question. The results were reported in three categories: 46% of respondents said they approved, 36% said they disapproved, and 18% said they had no opinion. Then the same question was asked of a separate sample of n = 881 people, all of whom had identified themselves as supporters of the “Tea Party” movement. The responses broke down like this in the same three categories:

Approve Disapprove No opinion
Tea Party Sample  352  432  97

(Data is adapted from the New York Times web site.) It seems from the numbers that “Tea Party” supporters may differ from the general population when it comes to their opinions concerning their Congressional Representatives. But is this a real difference, or could it just be a product of chance variation in the sampling process?

We did not discuss this in class, but a standard statistical method in this case would be to use a χ^{2}-test. The test statistic

X^{2} = ∑_{categories} \frac{(observed − expected)^{2}}{expected}

has approximately a χ^{2} distribution with degrees of freedom one less than the number of categories. For a test of H_{0} : the distributions are really the same, versus H_{a} :, we would reject H_{0} if X^{2} fell in an upper tail rejection region.

A) (5) If the “Tea Party” sample did follow the national sample percentages, how many responses would we expect in each category?

Solution: If the “Tea Partiers” were really the same as the general population in this regard, then the expected numbers of people in the Approve, Disapprove, and No opinion categories would be:

.46 × 881 ≐ 405.3, .36 × 881 ≐ 317.2, .18 × 881 ≐ 158.6.

B) (15) Carry out the χ^{2} test and interpret the results.

Solution: We compute the X^{2} statistic using the entries from the table above for the observed:

χ^{2} = \frac{(352 − 405.3)^{2}}{405.3} + \frac{(432 − 317.2)^{2}}{317.2} + \frac{(97 − 158.6)^{2}}{158.6} ≐ 72.5.

From the χ^{2} table with 2 degrees of freedom, we see that p < .005. This is extremely strong evidence that “Tea Partiers” have a different set of opinions on this question. (They are much less likely to approve, and also much less likely not to have an opinion!)